identify the product in the given reaction:

${\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}_2+\mathrm{NOCl}\to \mathrm{product}$

1. ${\mathrm{CH}}_3\mathrm{CHCl}·\mathrm{CH}2·\mathrm{NO}$

2. ${\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{NO}\right)·{\mathrm{CH}}_2\mathrm{Cl}$

3. ${\mathrm{CH}}_3\mathrm{CHCH}\left(\mathrm{Cl}\right)\left(\mathrm{NO}\right)$

4. ${\mathrm{CH}}_2\left(\mathrm{NO}\right)·{\mathrm{CH}}_2·{\mathrm{CH}}_2\mathrm{Cl}$

 3-methyl-2-pentene on reaction with HOCl gives:

1. 

2. 

3. 

4. 

"The negative part of the addendum adds on the carbon atom joined to the least number of hydrogen atoms." This statement is called as:

(1) Markownikoff's rule

(2) Peroxide effect

(3) Baeyer's strain theory

(4) Thiele's theory

Ease of abstraction of hydrogen is greater when attached to:

(1) 1$°$ carbon

(2) 2$°$ carbon

(3) 3$°$carbon

(4) neo carbon

Carbanion can undergo:

1. rearrangement

2. combination with cation

3. addition to a carbonyl group

4. all of the above are correct

The compound which reacts with HBr obeying Markownikoff's rule is:

1. $C{H}_2=C{H}_2$   

2. 

3. 

4. 

Arrange decreasing order of reactivity of given compounds with NBS/ ${\mathrm{CCl}}_4$/ hv:

a. ${\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{CH}}_3$

b. ${\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_3$

c. ${\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{CH}}_2-\mathrm{CH}={\mathrm{CH}}_2$

d. ${\mathrm{C}}_6{\mathrm{H}}_5-\underset{\underset{{\mathrm{CH}}_3}{|}}{\mathrm{C}}\mathrm{H}-\mathrm{CH}={\mathrm{CH}}_2$

Select the correct answer from the codes given below:

Codes :

1. d c a b

2. d c b a 

3. a b c d

4. a c b d

In the given reaction,

${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3\stackrel{\mathrm{Na}/{\mathrm{NH}}_3\left(\mathcal{l}\right)}{\to }\left[\mathrm{X}\right]$

[X] will be 

1. 1-phenylpropane

2. 1-phenylpropene

3. trans-1-phenyl propene

4. cis-1-phenylpropene

In the given reaction

$\underset{{\mathrm{N}}_2 \mathrm{gas}}{\overset{{\mathrm{CH}}_2{\mathrm{N}}_2/\mathrm{h\nu }}{\to }}$

The product will be:

1. 

2. 

3. Mixture of 1 and 2 

4. $C{H}_3-C{H}_2-C{H}_2-C{H}_2-C{H}_3$

In the given reaction,

$\stackrel{\mathrm{Br}}{\stackrel{|}{{\mathrm{CH}}_2}}-CH=C{H}_2\stackrel{NaN{H}_2/∆}{\to }\left[X\right]$

[X] will be:

(1) $C{H}_3-C\equiv CH$

(2) $H_{2}C=C\equiv C{H}_2$

(3) $C{H}_3-C\equiv \stackrel{⊝}{C}\stackrel{\oplus }{N}a$

(4) $H_{2}C=C\equiv \stackrel{⊝}{C}H\stackrel{\oplus }{N}a$